Regression analysis was applied between sales data (y in $1000s) and advertising data (x in $100s) and the following information was obtained. $$\begin{array} { l }
\widehat { y } = 12 + 1.8 x \
n = 17 \
\mathrm { SSR } = 225 \
\mathrm { SSE } = 75 \
\mathrm {~s} b _ { 1 } = .2683
\end{array}$$
Based on the above estimated regression equation, if advertising is $3000, then the point estimate for sales (in dollars) is
A) $66,000.
B) $5412.
C) $66.
D) $17,400.

Question

Regression analysis was applied between sales data (y in $1000s) and advertising data (x in $100s) and the following information was obtained.   $$\begin{array} { l } 
\widehat { y } = 12 + 1.8 x \
n = 17 \
\mathrm { SSR } = 225 \
\mathrm { SSE } = 75 \
\mathrm {~s} b _ { 1 } = .2683
\end{array}$$ 
Based on the above estimated regression equation, if advertising is $3000, then the point estimate for sales (in dollars) is
A) $66,000.
B) $5412.
C) $66.
D) $17,400.

Quizplus · Accepted Answer

The point estimate for sales is found by plugging the value of advertising into the estimated regression equation:$$\hat{y} = 12 + 1.8x$$$$\hat{y} = 12 + 1.8(3000)$$$$\hat{y} = 66,000$$Therefore, the point estimate for sales is $66,000.

Regression Analysis Was Applied Between Sales Data (Y in $1000s)